Rotation Matrix

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(0,0)

\theta

(x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)

Start with a general $2 \times 2$ matrix:

\begin{pmatrix} r & s \\ t & u \end{pmatrix}

It represents this rotation only if it $produces the same coordinates$ for every $x$ and $y$ .

This matrix must satisfy:

\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta - y \sin \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}

Expanding the multiplication on the left gives:

\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} x \cos \theta - y \sin \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}

By equating the corresponding components:

First row: $rx + sy = x \cos \theta - y \sin \theta$ .

For this equation to hold for all $x$ and $y$ , the coefficients of $x$ must be equal and the coefficients of $y$ must be equal. Thus, $r = \cos \theta$ and $s = -\sin \theta$ .
Second row: $tx + uy = x \sin \theta + y \cos \theta$ .

Similarly, $t = \sin \theta$ and .

Rotation Matrix about the Origin

Each $matrix column$ is the image of a unit vector. The horizontal unit vector rotates to $(\cos\theta,\sin\theta)$ . The vertical unit vector starts a quarter turn ahead, so its image is $(-\sin\theta,\cos\theta)$ . Use the same angle unit for sine and cosine:

R_\theta = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

Rotating about an Arbitrary Center

To rotate a point $(x,y)$ about an arbitrary point $P(a,b)$ by an angle $\theta$ , we $translate, rotate, then translate back$ in three steps:

Translate the point $(x,y)$ so that $P(a,b)$ becomes the origin: $(x-a, y-b)$ .
Rotate the translated point about the origin by $\theta$ using the matrix $R_\theta$ .
Translate the rotated point back by adding $(a,b)$ .

Centered Matrix Formula

This operation acts on the $vector relative to the center$ , not directly on the position vector from the origin:

\begin{aligned} \begin{pmatrix} x' \\ y' \end{pmatrix} &= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x-a \\ y-b \end{pmatrix} \\ &\quad{}+ \begin{pmatrix} a \\ b \end{pmatrix} \end{aligned}

When the point being rotated is the center itself, the relative vector is zero. The center therefore remains fixed:

R_\theta\left(\begin{pmatrix}a\\b\end{pmatrix}-\begin{pmatrix}a\\b\end{pmatrix}\right)+\begin{pmatrix}a\\b\end{pmatrix}=R_\theta\begin{pmatrix}0\\0\end{pmatrix}+\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}a\\b\end{pmatrix}

Finding a Specific Rotation Matrix

For $\theta = \frac{1}{4}\pi$ radians, or $45^\circ$ , both $trigonometric values$ equal $\frac{1}{\sqrt{2}}$ :

\begin{aligned} R_{\frac{\pi}{4}} &= \begin{pmatrix} \cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4}) \\ \sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4}) \end{pmatrix} \\ &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \end{aligned}

This matrix rotates every position vector by $45^\circ$ counter-clockwise.

For the point in the diagram, matrix multiplication gives:

\begin{pmatrix}1/\sqrt2&-1/\sqrt2\\1/\sqrt2&1/\sqrt2\end{pmatrix}\begin{pmatrix}2\\0\end{pmatrix}=\begin{pmatrix}2/\sqrt2\\2/\sqrt2\end{pmatrix}=\begin{pmatrix}\sqrt2\\\sqrt2\end{pmatrix}

Visualization of Rotating Point

(2,0)

by

45^\circ

about the Origin

Point

A(2,0)

is rotated by

45^\circ

to become

A'(\sqrt{2}, \sqrt{2})

.

\sqrt{2} \approx 1.414

.

Exercises

The problems move from building the rotation matrix to applying it. The last one rotates about $a center that is not the origin$ .

Determine the matrix associated with a rotation about the origin $O(0,0)$ by $\frac{1}{6}\pi$ radians.
Determine the image of point $P(4,2)$ if it is rotated about the origin $O(0,0)$ by $60^\circ$ .
Determine the image of point $Q(3,1)$ if it is rotated about the point $C(1,-2)$ by $90^\circ$ .

Solutions

Watch the $negative sign in the top-right entry$ of the rotation matrix. For an arbitrary center, subtract and add the center coordinates outside the rotation multiplication.

$Solution 1$

For $\theta = \frac{\pi}{6}$ , or $30^\circ$ , the sine and cosine are:

\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}

Substituting these values gives:

\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}

$Solution 2$

For point $P(4,2)$ and $\theta = 60^\circ$ , use $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$ .

\begin{aligned} \begin{pmatrix} x' \\ y' \end{pmatrix} &= \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} (4)(\frac{1}{2}) - (2)(\frac{\sqrt{3}}{2}) \\ (4)(\frac{\sqrt{3}}{2}) + (2)(\frac{1}{2}) \end{pmatrix} \\ &= \begin{pmatrix} 2 - \sqrt{3} \\ 2\sqrt{3} + 1 \end{pmatrix} \end{aligned}

The image is $P'(2-\sqrt{3}, 2\sqrt{3}+1)$ .

$Solution 3$

The point being rotated is $Q(3,1)$ , the center of rotation is $C(1,-2)$ , and $\theta = 90^\circ$ . Thus, $(a,b)=(1,-2)$ .

\cos 90^\circ = 0

The image is $Q'(-2,0)$ .

A quarter turn about a shifted center

Q rotates ninety degrees counterclockwise about C. Both radii keep the same length.

u = \cos \theta

Rotation Matrix

Matching Matrix Entries to Rotated Coordinates

Rotation Matrix

Matching Matrix Entries to Rotated Coordinates

Rotation Matrix about the Origin

Rotating about an Arbitrary Center

Centered Matrix Formula

Finding a Specific Rotation Matrix

Exercises

Solutions